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Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.

My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$, then do the same for $x^2-x+1$ and then use that along with the known remainder of $f$ divided by $x^2+x+1$ and $x^2-x+1$ to obtain $A$, $B$, $C$, $D$. However, $Ax^3$ is in the way so I don't know how to proceed nor do I have any other ideas to start with.

user562834
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  • Hint: $g=h_1 h_2$, $\gcd(h_1,h_2)=1$. Use the Chinese remainder theorem. – lhf Jun 22 '21 at 01:00
  • Do you know Chinese remainder theorem? If yes then use that to solve the q you would be able to solve it –  Jun 22 '21 at 04:00

5 Answers5

3

Using the Extended Euclidean Algorithm as described in this answer and adapted to polynomials yields $$ \begin{array}{r|r|r|r} \bbox[5px,border:2px solid #C00]{x^2+x+1}&\bbox[5px,border:2px solid #C00]{1}&0\\ \bbox[5px,border:2px solid #090]{x^2-x+1}&0&\bbox[5px,border:2px solid #090]{1}\\ 2x\phantom{{}+0}&1&-1&1\\ 1&\bbox[5px,border:2px solid #C00]{-\frac12x+\frac12}&\bbox[5px,border:2px solid #090]{\frac12x+\frac12}&\frac12x-\frac12\\ 0&x^2\phantom{\frac12}-x+\,1&-x^2\phantom{\frac12}-x-\,1&2x\phantom{{}+\frac12} \end{array}\tag1 $$ which says that $$ \overbrace{\color{#C00}{\left(x^2+x+1\right)\left(-\tfrac12x+\tfrac12\right)}}^{\large\frac12-\frac12x^3}+\overbrace{\color{#090}{\left(x^2-x+1\right)\left(\tfrac12x+\tfrac12\right)}}^{\large\frac12+\frac12x^3}=1\tag2 $$ Which, in turn says that $$ \frac12-\frac12x^3\equiv\left\{\begin{array}{} 0&\bmod x^2+x+1\\ 1&\bmod x^2-x+1 \end{array}\right.\tag3 $$ and $$ \frac12+\frac12x^3\equiv\left\{\begin{array}{} 1&\bmod x^2+x+1\\ 0&\bmod x^2-x+1 \end{array}\right.\tag4 $$ Therefore, the desired polynomial is $$ \begin{align} \scriptsize(-x+1)\overbrace{\left(\tfrac12+\tfrac12x^3\right)}^{\substack{1\bmod x^2+x+1\\0\bmod x^2-x+1}}+(3x+5)\overbrace{\left(\tfrac12-\tfrac12x^3\right)}^{\substack{0\bmod x^2+x+1\\1\bmod x^2-x+1}} &=3+x-2x^3-2x^4\\ &\equiv\bbox[5px,border:2px solid #CA0]{5+x+2x^2-2x^3}\bmod x^4+x^2+1\tag5 \end{align} $$

robjohn
  • 345,667
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This is where things go wrong:

My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$

The remainder inside the brackets can have a linear term as well, so you should have written it as $$ f(x) = (x^2 + x + 1)[(x^2 - x + 1)q(x) + Ax + (B-A)] + (C-B)x + (D-B+A). $$ And similarly, $$ f(x) = (x^2 - x + 1)[(x^2 - x + 1)q(x) + Ax + (B+A)] + (C+B)x + (D-B-A), $$ at which point you can solve for all four coefficients.

eyeballfrog
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Let the polynomial be $P(x)$. It is given that for some polynomials $Q(x),Q_1(x)$ $$P(x)=(x^2+x+1)Q(x)+1-x$$ $$P(x)=(x^2-x+1)Q_1(x)+3x+5$$ Now it is well known that $x^2+x+1$ has the zeros as $\omega,\omega^2$ and $x^2-x+1$ has the zeroes $-\omega,-\omega^2$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ So We get $$P(\omega)=1-\omega$$ $$P(\omega^2)=1-\omega^2$$ $$P(-\omega)=3\omega+5$$ $$P(-\omega^2)=-3\omega^2+5$$

Now let us assume for some polynomial $Q_2(x)$ we have $$P(x)=(x^4+x^2+1)Q_2(x)+Ax^3+Bx^2+Cx+D$$

Using the fact that $x^4+x^2+1$ has the zeros $\omega,\omega^2,-\omega,-\omega^2$ we get four linear equations as: $$\left[\begin{array}{cccc} 1 & \omega^{2} & \omega & 1 \\ 1 & \omega & \omega^{2} & 1 \\ -1 & \omega^{2} & -\omega & 1 \\ -1 & \omega & -\omega^{2} & 1 \end{array}\right]\left[\begin{array}{c} A \\ B \\ C \\ D \end{array}\right]=\left[\begin{array}{c} 1-\omega \\ 1-\omega^{2} \\ 5-3 \omega \\ 5-3 \omega^{2} \end{array}\right]$$

Now let us solve by Cramer's rule. The determinant of the matrix is $\Delta =12$, $\Delta_1=-24$, $\Delta_2=24$, $\Delta_3=12$ and $\Delta_4=60$. Thus the values of $A,B,C,D$ are $$\begin{aligned} &A=\frac{\Delta_{1}}{\Delta}=-2 \\ &B=\frac{\Delta_{2}}{\Delta}=2 \\ &C=\frac{\Delta_{3}}{\Delta}=1 \\ &D=\frac{\Delta_{4}}{\Delta}=5 \end{aligned}$$

So the required remainder is $$-2x^3+2x^2+x+5$$

Umesh shankar
  • 10,219
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Let $u=x^2+x+1,$ $v=x^2-x+1$ so that $uv=x^4+x^2+1.$ Let $P$ be the given polynomial. $$P=uq_1-x+1\tag1$$ $$P=vq_2+3x+5\tag2$$ $$P=uvq_3+R\tag3$$ There are polynomials $\lambda$ and $\mu$ such that $$\lambda u+\mu v=1\tag4$$ Suitable values are $\lambda={-x+1\over2}$ and $\mu= {x+1\over2} .$ By (1), (2) and (4), $$P=\lambda uP+\mu vP=\lambda uvq_2+(3x+5)\lambda u+\mu uvq_1+(-x+1)\mu v$$ Therefore, modulo $uv,$ $R$ is $$ (3x+5)\lambda u+(-x+1)\mu v\tag5 $$ Of course, we must reduce the degree in (5). $$(3x+5)\lambda=(3x+5){-x+1\over2}={1\over2}(-5x+8-3v)\tag6$$ $$(-x+1)\mu=(-x+1){x+1\over2}={1\over2}(x+2-u)\tag7$$ By (5), (6) and (7), $$R={1\over2}\big((-5x+8)(x^2+x+1)+(x+2)(x^2-x+1) \big)$$ $$= {1\over2}\big((-5x^3+3x^2+3x+8)+(x^3+x^2-x+2) \big)$$ $$\text{Thus,}\ \ \ R=-2x^3+2x^2+x+5. $$

David
  • 11
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Hint:

As the roots of $x^2+x+1=0$ are $w,w^2$ where $w$ is a complex cube root of unity,

the roots of $x^2-x+1=0$ are $-w,-w^2$

we can write $$f(x)$$

$$=p(x)(x^2+x+1)(x^2-x+1)+A(x-w)(x-w^2)(x+w)+B(x-w)(x-w^2)(x+w^2)+C(x-w^2)(x+w)(x+w^2)+D(x-w)(x+w)(x+w^2)$$

$$=p(x)(x^2+x+1)(x^2-x+1)+(x^2-x+1)(c(x-w^2)+d(x-w))+\cdots$$

$$-w+1=f(w)=(-2w)c(w-w^2)\implies2c=-w$$

Similarly, we can find $a,b,d$

lab bhattacharjee
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